What is the rate at which mechanical energy is dissipated?

[Rubidium]

1. Homework Statement
During a cold day, you can warm your hands by rubbing them together. Assume the coefficient of kinetic friction between your hands is 0.500, the normal force between your hands is 35.0 N, and that you rub them together at an average relative speed of 35.0 cm/s.
(a) What is the rate at which mechanical energy is dissipated?
(b) Assume further that the mass of each of your hands is 350 g, the specific heat of your hands is 4.00 kJ/(kg\cdotK), and that all the dissipated mechanical energy goes into increasing the temperature of your hands.
How long must you rub your hands together to produce a 5.00\circC increase in their temperature?




2. Homework Equations
(a) \DeltaE_{int}=Q_{in}+W_{on}
f_{k}=\mu_{k}F_{N}
(b) dE_{int}=c_{v}mdT




3. The Attempt at a Solution
(a) f_{k}=\mu_{k}F_{N}=(0.500)(35.0 N)=17.5 N \times0.35 m/s=6.125 J/s=6.13 W
(I know that part is correct.)
(b) dE_{int}=c_{v}mdT=(4.00 kJ/(kg\cdotK))(0.35 kg)(5.00\circC)=7.00 kJ
6.13 J/s / 0.007 J = 876 s
The correct answer is actually 38.1 min = 2286 s but I don't know what I am doing wrong. Please help.

 

[Dick]

You have two hands, so you will actually need 14kJ. And this "6.13 J/s / 0.007 J = 876 s" is nonsense, you've got different units on both sides! Find 14kJ/(6.13J/s).

 

[arunma]

You can differentiate equation (b) to get:

P = mc \frac{\Delta T}{\Delta t}

which can be rewritten as

\Delta T = mc\frac{\Delta T}{P}

Of couse you have to hands, so the actual formula should be:

\Delta t = 2mc\frac{\Delta T}{P}

If you plug in the numerical values, you'll find that you get the correct solution.

 

[Oerg]

the mass of both hands. You ahve to multiply the mass of one hand by 2. Also, 7kJ is 7000J.